
#91




Heh reminder to do that tomorrow.
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#92




T_T irresponsible^
Alright, I found out a couple very interesting things, but I still don't know WTF happened between my last two GIFs. First, I've verified completely that: Code:
1. cbrt(cos 87 + i sin 87) == cos 29 + i sin 29 2. cbrt(cos 87  i sin 87) == cbrt(cos 87 + i sin 87) == cos 29  i sin 29 I was not able to prove...that cbrt(64 cos 87 + 64i sin 87), is the same thing as 4 * cbrt(cos 87 + i sin 87) ... I used the complex numbers calculator I found earlier to check both expressions. They are different it seems, somehow?? cbrt(64 cos 87 + 64i sin 87) ~= (3.428669203 + 2.0601523i) 4(cos29 + i sin 29) ~= (3.49847884 + 1.9392385i) Again, my $100 TI84+ graphing calculator from Aug 31 found the cube root of (a + b) or that first line's approximation, which eventually verified into the sine of one degree! But when I factor cbrt(64) out of the radical, why I can't get the same expression...I'm sure I must have made some kind of ADHD assumption here that I will get to find out myself later. Well whatever. Still, as I've said I found a couple new useful things. My later code was actually for not the sine of one degree, but the cosine of thirtyone degrees. Code:
4sin(59) = [(1 + sqrt(3)i)(cos 29 + i sin 29)  (1  sqrt(3)i)(cos 29 + i sin 29)] ~= [1.7143346014 + 1.03007614982i  (1.7143346014 + 1.03007614982i)] ~= (1.7143346014 + 1.03007614982i  1.7143346014 + 1.03007614982i) ~= 3.4286692028 4 * sin(+59.) ~= +3.4286692028 sin(+59.) ~= +0.8571673007 cos(+31.) = sin(+59.) Code:
16sin(+1.) cbrt(64cos(87) + 64sin(87)i)(1 + sqrt(3)i) + cbrt(64cos(87)  64sin(87)i)(1  sqrt(3)i) (3.4286692 + 2.0601523i)(1 + 1.7320508i) + (3.428669203  2.0601523i)(1  1.7320508i) 0.13961924 + 7.9987815i  0.13961924  7.9987815i 0.2792384 sin(+1.) ~= 0.0174524 cbrt(64cos87 + 64sin(87)i) = 4 * cbrt(cos 87 + i sin 87) cbrt(64cos87 + 64sin(87)i) ~= (3.428669203 + 2.0601523i) 4(cos 29 + i sin 29) ~= (0.87461971 + 0.48480962i)
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http://theoatmeal.com/comics/cat_vs_internet Last edited by HatCat; 2nd November 2011 at 05:55 PM. 
#93




Shit lol, more reminders are in order.
What was I going to do next... cbrt(64 cos 87 + 64i sin 87) cbrt(64) * cbrt(cos 87 + i sin 87) Still confused by why in the hell I couldn't prove those were equal. Other than that I'm sure I was going to plug in the exact values for sin or cos, 29 or 87. It's very poorly documented online how to do denesting of nested cube roots, by comparison to all that's written about cube roots. Still I can't remember where exactly I left off, but I still have my 60 pages of inkstained history to look back at if I really need to XD.
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#94




Quote:

#95




...And then the segment AB said, "Oh, segment AC, how could you be without me? "
eww, no 1D 18+ material allowed *bans*
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#96




sorry about that lmao
I corrected the problem before, somehow. I'm too relieved to be bothered with what went wrong, but it was obviously something very skippy and inattentive. All it takes is setting up the original equation and, as anyone else would, multiplying. So, cbrt(64 cos 87 + 64i sin 87) == 4 cos 29 + 4i sin 29 .... Oh wait, probably what happened is that the online complex numbers calculator somehow interpreted differently. I factored out 4 to express as 4(cos 29 + i sin 29) but didn't have that issue this time. At any rate, still, using these statements I pasted into Notepad: Code:
cos 87 ~= 0.05233595624294383272211862960908 isin 87 ~= 0.99862953475457387378449205843944i cos 29 ~= 0.87461970713939580028463695866108 isin 29 ~= 0.48480962024633702907537962241578i http://www.solvemymath.com/online_ma...mber/index.php justifies this equality. Plugging (4 cos 29 + 4i sin 29) in for the cube root of (a + b), however, just resolves the system to a very fundamental and easy to derive statement for the sine of one degree caused by geometric derivation, which is worthless, because to finish evaluating you need to know the exact value of the sin or cos of 29 deg, which is 30  1 or 25 + 4, etc., all kinds of values we don't have the exact values of without them being in terms of the imaginary unit. For this reason I don't have a picture to post here yet, but I will later. I'll try to stay back in action tomorrow XD.
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#97




T_T kinda put this sort of thing off again, but so as to prevent the distraction of the previous matter from the future work, here's an image showing what I meant.
OpenOffice.org's math application has this shitty way of handling the font...won't even export a PDF with a smooth render of the algebra if there's more than one line or whatever. Halfassed support. Anyway, I haven't thoroughly checked everything there. All I know is that I proved that cbrt(a + b/* == 64 cos 87 + 64i sin 87 */) is equivalent to (4 cos 29 + 4i sin 29). Plugged that in and simplified from there. Maybe I glitched somewhere along the lines, maybe not. I'll check later. Or actually I should be able to check easily now so just do that now.... If the last statement is true: Code:
sin(1) ?= 0.5i[(sin(29)  cos(29)(sqrt(3))] [sin(1)]^2 ?= 0.25[(sin(29)  cos(29)(sqrt(3))]^2 [sin(1)]^2 ?= 0.25[cos(29)(sqrt(3))  (sin(29)]^2 [sin(1)]^2 != +0.26526421860705461202026885588608 /* nobrainer, but just for reference, what I got for squared sine of 1 [deg] is ~ +0.00030458649045213499687827997803515 */
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#98




XD the moment I posted that I instantly saw it was between the fourth and fifth lines. All the sign negations and switching expressions around in the form of (a  b) as opposed to (b + a), just to remove the amount of text characters I had to type in the original text code to generate the image, proved to be quite hazardous. I was working way too fast when I wrote that GIF lmao. Like I said fix later
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http://theoatmeal.com/comics/cat_vs_internet Last edited by HatCat; 2nd December 2011 at 08:49 AM. 
#99




What's funny about this is that by making that additive inversion error between the fourth and fifth lines, I accidentally defined the sine of 31 degrees.
While by correcting the issue from top to bottom, I defined instead the sine of 59 degrees. I verified this in the topmost statement as well using the complex numbers calculator. (Yes I can see the white spots before lol; I lost part of my old software archive and was too lazy to optimize the GIF correctly.) Obviously I lost track of the correct expression I'm supposed to be using or something XD, so I really need to rewind. But hey, who knows, maybe the sine of fiftynine degrees (or cos 31) and the sine of thirtyone degrees (or cos 59) will come in handy in a later process, although honestly I doubt that. Interesting though...lots of patterns. The easiest one to notice is sin 30 [deg] = 1:2 (as in, onehalf times cosine of 29 degrees) and that onehalf the square root of three (the coefficient of sin 29), is the sine of 60 degrees. This is nothing new or special...just a demonstration of a wellknown theorem. sin(m + n) = sin(m) • cos(n) + cos(m) • sin(n) sin(59) = sin(30 + 29) = sin(30) * cos(29) + cos(30) * sin(29) Still, I must have picked the wrong image in this thread or something...how in the hell was I doing sin(59) all along XD.
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#100




Figured it out...the reason my original verification of my rewritten expressions for the sine of one degree maintained validity was because of the solution I was then using to simplify the cube root expression, in terms of a and b, as opposed to the new one I calculated using de Moivre's theorem. It turns out that that was, in fact, a valid cube root of (a + b), but it was part of what has turned out to be a multiplicity of valid solutions. Cube root functions return only one value, as a function must always, but it seems that there is a cube root relation when the operand is a complex number, in which case there can be multiple values outside of the real number line for solution when solution • solution • solution == (a + b).
I went back to the value I was using previously, which I found using my graphing calculator from August, and I found that this decimal approximation was slightly but significantly different than the valid but, still, other expression whose decimal approximation was enough to still also cube into the original binomial, the sum of a and b, as well. I instead used the previous and found the inverse sine and cosine function evaluations. Code:
sin{1} = {(1 + sqrt{3}i)(4 cos{31} + 4 sin{31}i) + (1  sqrt{3}i)(4 cos{31}  4 sin{31}i)} over 16 newline sin{1} = {(1 + sqrt{3}i)(cos{31} + sin{31}i) + (1  sqrt{3}i)(cos{31}  sin{31}i)} over 4 newline sin{1} = {cos{31} + sin{31}i + sqrt{3}cos{31}i  sqrt{3}sin{31} + cos{31}  sin{31}i  sqrt{3}cos{31}i  sqrt{3}sin{31}} over {4} newline sin{1} = {2 cos{31}  2 sqrt{3}sin{31}} over 4 newline sin{1} = (1 over 2) cos{31}  (sqrt{3} over 2) sin{31} newline sin{1} = (sqrt{3} over 2) sin{31}  (1 over 2) cos{31} Now, as opposed to before, this time we arrive at this application of the form: sin(1) = sin(31  30) = sin(31) * cos(30)  cos(31) * sin(30) I can't imagine any way to write as sin(m + n) because it would have to be nonwholenumbers or something redundant like sin(0 + 1) = sin(0) + cos(1) + cos(0) * sin(1), which tells us nothing. Anyway, now that I've gotten that distraction out of the way, I can either return to where I was originally at and denest the cube roots in terms of sin or cos 87 (which is divisible by 3 and not defined in terms of the imaginary unit on most websites today, easier in that aspect) or, possibly, use the definition for the sine or cosine of thirtyone degrees (as discovered in this little adventure here XD) and modify the complex number to be of the form (a + 0i) so that the imaginary unit does not interfere with the rest of the process. If I can do the latter then that will be loads easier XD.
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